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If $x,y>1$ and $(\ln x)^2+(\ln y)^2=\ln x^2+\ln y^2.$ then maximum value of $x^{\ln y}$

we can write $(\ln x)^2-\ln (x^2) = -((\ln y)^2-\ln (y^2))$ and assuming $x\geq y$

now let $f(x)=(\ln x)^2-\ln (x^2) $

$\displaystyle f'(x)=2 \ln x \cdot \frac{1}{x}-\frac{1}{x^2}\cdot 2x =\frac{2(\ln x-1)}{x}$

and $\displaystyle f''(x) = \frac{2(2-\ln x)}{x^2}$

wan,t be able to go further, could some help me with this, thanks

  • 0
    try $x=2,y=1$ then we have $$(\ln(2))^2=\ln(4)$$ and this is not the same – 2017-01-13
  • 0
    Setting $x = e^u$ and $y=e^v$, we have to maximize $f(u,v) :=\exp(u (\sqrt{1 + 2u - u^2} - 1))$, which is just basic calculus. – 2017-01-13

2 Answers 2

1

Hint

$$k=x^{\ln y} \Leftrightarrow \ln k=\ln x \cdot \ln y$$

So we can analyse $\ln x \cdot \ln y$, once $\ln$ is increasing.

$$(\ln x)^2+(\ln y)^2=2\ln x+2\ln y \Leftrightarrow (\ln x-1)^2+(\ln y -1)^2=2$$

Call, $p=\ln x$ and $q=\ln y$ then you have to find the maximum of $$p\cdot q$$ such that $$(p-1)^2+(q -1)^2=2\quad (1)$$

write $f(p)=pq$ then $f'(p)=q+pq'$ and by implicity derivation at $(1)$

$$p-1+q'(q-1)=0 \rightarrow q'=\frac{1-p}{q-1} \quad (2)$$

so, $f'(p)=0 \rightarrow pq'=-q\quad (3)$

Now you can find critical points $p$ and $q$ .

Remember that $x>1 \rightarrow p>0$.

Can you finish using $(1)$, $(2)$ and $(3)$?

2

Hint Try making perfect squares, you'll get a circle. Then use parametric substitution and find the value.